A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 2.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring

As the box compresses the spring, the spring performs

-1/2 (85 N/m) (0.065 m)² ≈ -0.18 J

of work on the box. By the work energy theorem, the total work performed on the box (which is done only by the spring since there's no friction) is equal to the change in the box's kinetic energy. At full compression, the box has zero instantaneous speed, so

W = ∆K   ==>   -0.18 J = 0 - 1/2 (2.5 kg) v ²

where v is the box's speed when it first comes into contact with the spring. Solve for v :

v ² ≈ 0.14 m²/s²   ==>   v ≈ 0.38 m/s

d is the correct answer